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Obtain the expression for electric field due to an infinitely long charged wire . |
Answer» Solution :Electric field DUE to an INFINITELY long charged wire: Consider an infinitely long straight wire having uniform linear charge density `lambda`. Let P be a point located at a perpendicular distance r from the wire. The electric at the point P can be found using Gauss law . We choose two small charge element `A_(1)` and `A_(2)` on the wire which are at equal distance from the point P. The resultant electric field due to these two charge elements points radially away from the charged wire and the magnitude of electric field is same at all points on the circle of radius r. From this property. We can infer that the charged wire possesses a cylindrical symmetry.  Let us choose a cylindrical Gaussian surface of radius r and length L . The total electric FLUX in this closed surface is `Phi_(E) =oint vecE . d VECA` `Phi_(E)=oint underset("Curved surface ")(oint)vecE.dvecA+ underset("Top surface")intvecE.dvecA+underset("Bottom surfac")int vecE.dvecA` It is seen that for the curved surface `vecE` is perpendicular to `vecA` and `vecE.dvecA` = 0 Substituting these value in the equation (2) and applying Gauss law to the cylindrical surfaces we hav `Phi_(E)= underset("Curved surface ")(int) EdA = (Q_("encl"))/(epsilon_(0))` SINCE the magnitude of the electric field for the entire curved surface is constant E is taken out of the integration and `Q_("encl") ` given by `Q_("encl") = lambdaL.` `E underset("Curved surface ")(int) dA = (lambdaL)/(epsilon_(0))`  Here `Phi_(E) = underset("Curved surface ")(int) dA ` = total area of the curved surface = `2piL` . Substituting this inequation (4) We get `E.2 pirL = (lambdaL)/(epsilon_(0))( or ) E =(1)/(2piepsilon_(0)) (lambda)/(r)` In vector form `vecE = (1)/(2piepsilon_(0))(lambda)/(r) hatr ` The electric field due to the infinite charged wire depends on `(1)/(r) ` rather than `(1)/(r^(2))` for a point charge . Equation (6) indicates that the electric field ios always along the perpendicular direction ( r) to wire . In fact if `lambda gt ` 0 then `vecE ` points perpendicular outward `(hatr)` from the wire and if `lambda lt 0` then `vecE` points perpendicular inward `(-hatr)` . |
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